Find Amp of Circuit with Ohm's Law

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Discussion Overview

The discussion revolves around calculating the current (amps) in a circuit using Ohm's Law, with a specific focus on a circuit made of copper with given resistivity and voltage. Participants explore the relationship between voltage, resistance, and current, while addressing the importance of units and additional factors affecting the calculation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests using Ohm's Law, stating that current equals voltage divided by resistance.
  • Another participant emphasizes the importance of correct terminology and units, clarifying the definitions of current, voltage, and resistance.
  • A different participant notes the necessity of including the source resistance of the voltage supply in the calculations.
  • One participant points out the need to convert resistivity to resistance, providing a formula that involves the length and cross-sectional area of the conductor.
  • Another participant calculates the current based on the provided resistivity and voltage, resulting in a very high current value, while cautioning about the realism of such a large current in practical applications.

Areas of Agreement / Disagreement

Participants generally agree on the application of Ohm's Law but express differing views on the importance of additional factors such as source resistance and the conversion from resistivity to resistance. The discussion remains unresolved regarding the practical implications of the calculated current.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the circuit components, the specific conditions under which the calculations apply, and the need for additional parameters such as length and cross-sectional area for accurate resistance calculations.

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Hi,
I'm trying to find the amp of a circuit. Okay I know a circuit has a resistivity, So for example if my circuit is made out of cooper which has a resistivity of 1.68 * 10 negative power of 8 and a voltage of 9 volts.So do I just use ohms law Volt / Resistant = Amp?
 
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Yes, use Ohm's Law. Current equals Voltage divided by resistance. For a great explanation and built-in calculator, see:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html

By the way, it's important to get your terminology and units correct

Current (I): units of Amperes
Voltage (E): units of Volts
Resistance (R): units of Ohms
 
Pin Head said:
Hi,
I'm trying to find the amp of a circuit. Okay I know a circuit has a resistivity, So for example if my circuit is made out of cooper which has a resistivity of 1.68 * 10 negative power of 8 and a voltage of 9 volts.So do I just use ohms law Volt / Resistant = Amp?

Be sure to carry units along in your calculations -- it's important to have the units of the resistivity in order to get the correct answer for the resistance.

In this case, you would need to be sure to include the source resistance of your "9V" source. When you put a heavy load on a voltage source, its internal resistance needs to be part of your calculation (using Ohm's Law, yes).
 
Pin Head said:
Hi,
I'm trying to find the amp of a circuit. Okay I know a circuit has a resistivity, So for example if my circuit is made out of cooper which has a resistivity of 1.68 * 10 negative power of 8 and a voltage of 9 volts.So do I just use ohms law Volt / Resistant = Amp?

But first you will need to convert from resistivity to resistance.

For this, you need the formula:
140b5b8782cd1e72fa3785af25e1b394.png


Where
R = resistance of the metal sample.
ρ is the resistivity (in ohm -meters).
L is the length of the sample in the direction of current flow (in meters).
A is the cross sectional area of the sample (in square meters).
 
Last edited:


Hi there,

Yes, you are correct. Ohm's Law states that the current (in amps) flowing through a circuit is equal to the voltage (in volts) divided by the resistance (in ohms). So in your example, the current would be 9 volts divided by 1.68 * 10^-8 ohms, which equals 5.36 * 10^8 amps. Keep in mind that this is a very large amount of current and may not be realistic for a typical circuit. It's important to also consider other factors, such as the size and capacity of the circuit components, to ensure safe and efficient operation. Hope this helps!
 

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